Answer:
Precipitation Reactions
They contain two aqueous reactants, one aqueous product, and one solid product. In this reaction, two soluble products, Pb(NO3)2 and KI, combine to form one soluble product, KNO3, and one insoluble product, PbI2. This is a precipitation reaction, and PbI2 is the precipitate.
Answer is: because weak acids do not dissociate completely.
The strength of an Arrhenius
acid determines percentage of ionization of acid and the number of H⁺ ions formed. <span>
Strong acids completely ionize in water and give large amount ofhydrogen ions (H</span>⁺), so we use only one arrow, because reaction goes in one direction and there no molecules of acid in solution.
For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).
<span>
Weak acid partially ionize in water
and give only a few hydrogen ions (H</span>⁺), in the solution there molecules of acid and ions.
For example cyanide acid: HCN(aq) ⇄ H⁺(aq)
+ CN⁻(aq).
Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
